Chem 455 Quiz 1
8 March 2006
Name: ___________________
Question 1 (15 Points)
Consider the molecular orbital diagram below for an atom with an s and a p orbital with 2 atoms
with p orbital contributions. The molecular orbitals shown have a total of 12 electrons.
Please provide the following:
a. Place the 12 electrons in the molecular orbitals representing the ground state (see below)
b. Show transitions that correspond to
I. 
II. 
III. n
IV. n
c. Rank the transitions in energy from highest to lowest and provide a reasonable
wavelength absorbance range for each transition
I. 120-130 nm n150 nm to 250 nm),  and nabout the same
(200 to 700 nm)
n
*
n




*
1
Question 2 (15 Points)
Consider the following representative energy level data for Cr shown below.
a. Which transitions involving the 3d54s ground state would require a photon with a
wavelength between 400 nm and 500 nm? (See highlight in table)
b. Provide the energy (in J or eV) for one of the transitions involving the 3d54s ground state
(see data in table, solved using the transition wavelength in meters found from the
wavenumbers)
c. What is the energy difference (in J or eV) between the 3d5(6S)4p and 3d4(5D)4s4p(1P°).
Note: 2 possible levels for 3d5(6S)4p; 23305.01 cm -1 and 26787.50 cm-1
If 3d5(6S)4p energy is 23305.01 cm-1 and 3d4(5D)4s4p(1P°) is 40906.46 cm-1
The wavelength difference is 1/(40906.46 – 23305.01) cm = 5.68E-7 m
E=6.626E-34 Js*3E8ms-1/5.68E-7 m = 3.50E-19 J = 2.18 eV
Or
3d5(6S)4p energy is 23305.01cm-1; wavelength of 4.29E-7 m
3d4(5D)4s4p(1P°) is 40906.46 cm-1; wavelength of 2.44E-7m

E= hc(1/1-1/2)
E=6.626E-34 Js*3E8ms-1 (1/2.44E-7 –1/4.29E-7) = 3.51E-19 J = 2.19 eV
If 3d5(6S)4p energy is 26787.50 cm-1 and 3d4(5D)4s4p(1P°) is 40906.46 cm-1
The wavelength difference is 1/(40906.46 - 26787.50) cm = 7.08E-7 m
E=6.626E-34 Js*3E8ms-1/7.08E-7 m = 2.81E-19 J = 1.75 eV
Or
3d5(6S)4p energy is 26787.50 cm-1; wavelength of 3.72E-7 m
3d4(5D)4s4p(1P°) is 40906.46 cm-1; wavelength of 2.44E-7m

E= hc(1/1-1/2)
E=6.626E-34 Js*3E8ms-1 (1/2.44E-7 – 1/3.72E-7) = 2.81E-19 J = 1.75 eV
d. The ground state for Cr is 3d5(6S)4s. Why is this the ground state and not 3d44s2?
The d orbital is half-filled with 5 electrons, stabilizing the configuration over the pairing of
electrons in the 4s orbital
e. Why are there different energy levels for the 3d54s state? (i.e., 3d5(4G)4s is different from
3d5(4D)4s and both are different from the 6S ground state).
The different energy levels arise from the difference in electron configuration in the ground
state. The five electrons in the 3d orbitals are arranged so each occupies a separate orbital in
the ground state, hence the 6S term. The d orbitals are split, with some in a lower energy state
and some in a high energy state. An electron can be excited to the higher energy d orbital.
This results in an electron pair forming in the excited state with an accompanying decrease in
spin as represented by the
f. How many unpaired electrons are in the 3d orbital of the state 3d5(2I)4s?
2S+1=2, S=1/2, one unpaired electron.
2
Hint on energies. The energies are given in wavenumbers. Wavenumbers are the number of
waves in a given length and are also the reciprocal of the wavelength.
Configuration
3d5(6S)4s
3d44s2
Wavenumber cm-1
wavelength
(m)
GS transition
Energy (J)
GS transition Energy
(eV)
0.00
7750.78
1.29E-06
1.54E-19
9.61E-01
3d5(4G)4s
20517.40
4.87E-07
4.08E-19
2.54E+00
3d5(4P)4s
21840.84
4.58E-07
4.34E-19
2.71E+00
3d5(6S)4p
23305.01
4.29E-07
4.63E-19
2.89E+00
3d5(4D)4s
24277.06
4.12E-07
4.83E-19
3.01E+00
3d4(5D)4s4p(3P°)
25206.02
3.97E-07
5.01E-19
3.13E+00
3d5(6S)4p
26787.50
3.73E-07
5.32E-19
3.32E+00
3d4(5D)4s4p(3P°)
27300.19
3.66E-07
5.43E-19
3.39E+00
3d5(2D3)4s
31009.00
3.22E-07
6.16E-19
3.85E+00
3d5(2F1)4s
33040.10
3.03E-07
6.57E-19
4.10E+00
3d5(2I)4s
33762.74
2.96E-07
6.71E-19
4.19E+00
3d5(2H)4s
35870.53
2.79E-07
7.13E-19
4.45E+00
3d4(5D)4s4p(1P°)
40906.46
2.44E-07
8.13E-19
5.07E+00
Question 3 (10 Points)
Compare and contrast atomic absorption spectroscopy and atomic emission spectroscopy.
Both methods use atomization of a sample and hence determine the concentration of elements.
For absorption light of a defined wavelength is passed through a sample and the absorption of the
light is determined. The absorption is defined by the electron energy transitions for a given
element and is specific for the given element. The concentration is proportional to the absorbed
light. In atomic emission the element is excited. A rapid relaxation accompanied by emission of
UV or visible light is used to identify the element. The intensity of the emitted photon is
proportional to the element concentration.
Question 4 (10 Points)
Please provide information on possible interference during measurements for the following
methods.
a. Atomic absorption spectrometry
Atomic absorption interferences can be divided into chemical and spectral interferences.
Spectral interferences can arise from sample spectra overlap and scattering. Spectra overlap
for different elements in a sample are not very common and can be correct by selection of a
different transition. Scattering is due to formation of oxides in the flame during
measurement. Chemical interference is more common and stems from the formation of
3
compounds with low volatility, dissociation equilibria, or ionization equilibria. Additives
such as EDTA can be used to decrease volatility by removal of the compounds.
b. Atomic emission spectrometry
The main interference from AES comes from emission lines. In particular, Ar lines from the
torch can cause interference.
c. Atomic mass spectrometry
Interferences from mass spectrometry stem from having the same m/z for the analytes in a
plasma. If the charge on compounds varies, then m/z could represent different masses
with different charges. Isobaric interferences from elements with isotopes of similar mass
can occur. Examples include 113,115In with 113Cd and 115Sn; 40Ar with 40K; and oxides of
Ti (46-50Ti forming masses with 62-66) interfering with 62Ni, 63Cu, 64Zn, 65Cu, and 66Zn.
Other polyatomic interferences include dimers (O2, N2, and Ar2), metal hydroxides
(additional 17 mass units above metal isotope), and compounds listed below:
Interference
Analyte
38
Ar1H
39
40
Ar
40
Cl16O
51
35
K
Ca
40
Ar12C
52
23
Na40Ar
63
40
Ar35Cl
75
V
Cr
Cu
As
Question 5 (10 Points)
Compare and contrast different mass analyzers used in mass spectrometry.
 Quadrupole
A mass analyzer that has 4 electrodes; two of which are paired. Alternation of potential is
used to maintain the selection of a given m/z ratio. This selects a single m/z and acts as a
filter.
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 Time of flight
A sample is ionized and enters a drift tube with the same kinetic energy. The ion velocity
is inversely proportional to the mass, resulting in separation of ions based on mass. The
separation follows
m
t
 2eEl s ( ) 2
z
l
m= mass of analyte ion
z = charge on analyte ion
E = extraction field
t= time-of-flight of ion
ls = length of the source
l = length of the field-free drift region
e = electronic charge (1.6022E-19 C)
 Double focusing
This method uses a magnetic field to separate ions as a function of their momentum.
Electrostatic focuses ions in a narrow kinetic energy range. Heaviest ions are the least
perturbed.
Question 6 (15 Points)
a. Describe the electron transition processes involved in x-ray generation?
An electron is removed from a shell near the nucleus, most likely a K or L shell. An electron
in a shell further away from the nucleus occupies the vacated orbital. This transition is
accompanied by the emission of a photon that corresponds to the transition energy and is the
x-ray.
5
You have the following x-ray data
Element Z
Rh
Sn
Rh
Nd
Tb
45
50
55
60
65
K (keV)
20.2161
25.2713
30.9728
37.3610
44.4816
K(keV)
22.7236
28.4860
34.9869
42.2710
50.3820
You wish to determine the element in a sample measured by x-ray emission via fluorescence. A
K x-ray of 40.1181 keV is measured.
b. What is the element that produces the 40.1 keV K x-ray?
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To find the element from the x-ray one can use the relationship is from Moseley ( Z  fn(  )
and  is related to energy, Z  fn( E ) and the relationship is linear.
70
y = 3.6945 + 9.2043x R= 0.99995
65
Z
Z
60
55
50
45
40
4
4.5
5
5.5
6
6.5
7
Sqrt K alpha
The relationship for K is Z  9.2043 K   3.6945
With K=40.1181 keV, K   6.3339
Z=9.2043*6.3339+3.6945 = 61.99 = 62=Sm
For the K data above,
K   6.333 =0.110Z-0.315; 0.110Z=6.648=0.110Z; Z=
c. What is the energy of the K x-ray from the element that produced the 40.1 keV K x-ray?
Based on the ratios for K/Kfrom the data above, one get K/K=0.883. If the K for Sm is
41.1181 keV then the K is 45.43 keV.
From the linear regression one gets
7
Z
Data 1
70
y = 4.1702 + 8.5807x R= 0.99996
65
Z
60
55
50
45
40
4.5
5
5.5
6
6.5
7
7.5
Sqrt K beta
Solve for Z=62
K   6.7395 , K = 45.42 keV
The actual value is 45.413 keV
d. Which more intense, the K x-ray or the K x-ray?
The K x-ray is more intense. The LK electron transition has a higher probability than the
MK transition, hence the K x-ray is dominant.
Question 7 (10 Points)
Describe the theory and concepts behind molar absorptivity
Molar absorptivities range from 0 to 1E5. The resulting value for the molar absorptivity can be
thought to express two terms, a transition probability and a target cross section.
Question 8 (15 Points)
8
Two different Cr species CrL and CrL’, coexist in the same solution. You have prepared four
solution, each contain CrL and CrL’ but at different concentrations. The concentration of each
species varies for your four samples according to the table below. You place a sample of each
solution containing the CrL and CrL’mixture in a 10 mm x 10 mm cuvette and measure the
percent transmission at 610 nm.
What is the molar absorptivity for CrL and CrL’?
Solution # [CrL] M
[CrL'] M
% transmission
1
5.00E-05
2.00E-04
13.72
2
1.00E-04
1.50E-04
18.84
3
1.50E-04
1.00E-04
25.85
4
2.00E-04
5.00E-05
35.48
First one must convert the % transmission to absorbance using A=-log(T)
[CrL] M
[CrL'] M
% transmission
A
5.00E-05
2.00E-04
13.72 0.8625
1.00E-04
1.50E-04
18.84 0.725
1.50E-04
1.00E-04
25.85 0.5875
2.00E-04
5.00E-05
35.48
0.45
The total absorbance equals the contribution from each species. Beer’s law is A=bc, with b = 1
cm and one needs to consider the contribution from each species.
At=bc1+bc2
Use the data to solve for one component of the molar absorptivity.
0.8625=5E-5+2E-4
0.725=1E-4+1.5E-4
Normalize for  to solve forby multiplying the top equation by 2
1.725=1E-4+4E-4
-[0.725=1E-4+1.5E-4]
1.000=2.5E-4
1/2.5E-4 = 4000 cm-1moles-1L =for CrL’
Insert this an equation and solve for 1
0.725=1E-4+4000*1.5E-4
0.125=1E-4
1250 cm-1moles-1L = 1 for CrL
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